A Styrofoam bucket of negligible mass contains 1.60 kg of water and 0.425 kg of ice. More ice, from a refrigerator at -14.8 degree C, is added to the mixture in the bucket, and when thermal equilibrium has been reached, the total mass of ice in the bucket is 0.768 kg.
Assuming no heat exchange with the surroundings, what mass of ice was added?
m=??kg
The mixture of 1.6 kg of water and of 0.425 kg of ice must be at 0oC.
Mass m(new ice) at -14.8oC is added.
Q(of ice) = Q(freeze water)
m(new ice) • c•4180•(0 – (-14.8)) = m(freeze water) •r,
where r = 335000 J/kg is the heat of fusion.
c = 4180 J/kg•oC is heat capacity.
m(new ice) • 61864 = m(freeze water) •335000,
m(freeze water) = 0.185• m(new ice),
Mass of ice at the end = 0.768 kg.
So
0.768 kg = 0.425 kg + m(new ice) + m(freeze water),
0.768 kg = 0.425 kg + m(new ice) + 0.185• m(new ice),
(0.768 - 0.425) kg = 1.185• m(new ice),
m(new ice) = (0.768 - 0.425)/1.185 = 0.289 kg.
To find the mass of ice that was added, we can make use of the principle of conservation of energy. Since there is no heat exchange with the surroundings, the total amount of heat in the system remains constant.
First, let's determine the initial heat in the system. We know that the specific heat capacity of water is 4.18 J/g°C and the specific heat capacity of ice is 2.09 J/g°C.
The initial heat (Q1) in the system can be calculated as follows:
Q1 = mass of water * specific heat capacity of water * change in temperature of water
= 1.60 kg * 4.18 J/g°C * (0°C - 0°C)
= 0 J
Next, let's determine the final heat (Q2) in the system. The final temperature is not provided, but since the system has reached thermal equilibrium, we know that the final temperature of the mixture will be 0°C.
Q2 = (mass of water + mass of added ice) * specific heat capacity of water * change in temperature of mixture
= (1.60 kg + mass of added ice) * 4.18 J/g°C * (0°C - (-14.8°C))
Since the total amount of heat in the system is conserved, Q1 = Q2:
0 J = (1.60 kg + mass of added ice) * 4.18 J/g°C * (-14.8°C)
Simplifying the equation:
0 = -62.528 kg/J * (1.60 kg + mass of added ice)
0 = -99.648 + (-62.528 * mass of added ice)
62.528 * mass of added ice = -99.648
mass of added ice = -99.648 kg / 62.528 (kg/J)
mass of added ice ≈ -1.59 kg
The negative sign indicates an inconsistency in the given values or calculations, as mass cannot be negative. Please review the problem and recheck the values to find any discrepancies.
To find the mass of ice added to the bucket, we can use the principle of conservation of energy.
The heat lost by the ice being added (Q lost) will be equal to the heat gained by the ice and water in the bucket (Q gained). We can calculate these heats using the equation:
Q = mcΔT
Where:
- Q is the heat gained or lost
- m is the mass
- c is the specific heat capacity
- ΔT is the change in temperature
First, let's find the heat gained by the ice and water in the bucket. The original mass of ice and water in the bucket is 1.60 kg + 0.425 kg = 2.025 kg.
ΔT for the ice and water in the bucket can be calculated as the difference between the initial temperature of the ice in the bucket (-14.8°C) and the final temperature of thermal equilibrium. Since there is no heat exchange with the surroundings, the final temperature would be 0°C (the melting point of ice).
ΔT = 0°C - (-14.8°C) = 14.8°C
Using the specific heat capacity of water (c = 4.18 J/g·°C), we convert the mass of the ice and water in the bucket to grams.
m_ice_water = 2.025 kg * 1000 = 2025 g
Now, we can calculate the heat gained by the ice and water in the bucket.
Q gained = m_ice_water * c * ΔT
Q gained = 2025 g * 4.18 J/g·°C * 14.8°C
Next, let's calculate the heat lost by the ice being added to the bucket. We don't know the mass of the ice being added (m_ice_added), so we leave it as a variable.
The final mass of ice in the bucket is 0.768 kg converted to grams.
m_final_ice = 0.768 kg * 1000 = 768 g
The initial temperature of the ice being added is -14.8°C, and the final temperature at thermal equilibrium is 0°C.
ΔT_ice_added = 0°C - (-14.8°C) = 14.8°C
Using the specific heat capacity of ice (c = 2.09 J/g·°C), we can calculate the heat lost by the ice being added.
Q lost = m_ice_added * c * ΔT_ice_added
Now, we equate Q gained and Q lost:
Q gained = Q lost
m_ice_water * c * ΔT = m_ice_added * c * ΔT_ice_added
2025 g * 4.18 J/g·°C * 14.8°C = m_ice_added * 2.09 J/g·°C * 14.8°C
Simplifying the equation:
2025 g * 4.18 = m_ice_added * 2.09
Solving for m_ice_added:
m_ice_added = (2025 g * 4.18) / 2.09
m_ice_added = 4055.7 g
Finally, we convert the mass from grams to kilograms:
m_ice_added = 4055.7 g / 1000 = 4.0567 kg
Therefore, the mass of ice added to the bucket is approximately 4.06 kg.